Falacer comments on What should a Bayesian do given probability of proving X vs. of disproving X? - Less Wrong

0 Post author: PhilGoetz 07 June 2014 06:40PM

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Comment author: Falacer 07 June 2014 11:00:22PM *  0 points [-]

Goldbach's conjecture is "Every even integer above four is the sum of two primes," surely?

Also, Gödel's incompleteness theorem states that there are theorems which are true but non-provable, so you get something like:

P(X) = (P(will be proven(X)) + P(is true but unprovable(X))) / (P(will be proven(X)) + P(will be disproven(X)) + P(is true but unprovable(X)))

Is there a reason to suspect that a counterexample wouldn't be a very large number that hasn't been considered yet? Consider sublime numbers: the first (12) is a number which will be checked by any search process, but there is another which has 76 digits and, I would suspect, could be missed by some searches.

Comment author: DanielLC 08 June 2014 12:04:14AM 0 points [-]

Goldbach's conjecture is ...

Fixed. Thanks.

P(X) = (P(will be proven(X)) + P(is true but unprovable(X))) / (P(will be proven(X)) + P(will be disproven(X)) + P(is true but unprovable(X)))

It still only works on assumptions that might generally make good approximations, but aren't necessarily true.

More to the point, Phil was suggesting that something is false on the basis that it would be difficult to prove if true, but easy to prove if false. In other words, he was using it specifically because that example was one where the implicit assumption in his approximation, that the relative values of P(will be proven(X)) to P(will be disproven(X)) are about the same as P(is true but will not be proven(X)) to P(is false but will not be disproven(X)), was particularly far from the truth.

Is there a reason to suspect that a counterexample wouldn't be a very large number that hasn't been considered yet?

Suppose they'll check every number up to n. Suppose also that we're using a logarithmic prior as to where the first counterexample is. Suppose also we'll eventually find it. It has the same probability of being from one to sqrt(n) as from sqrt(n) to n. This means that if m, the number of numbers we've already checked, is more than sqrt(n), we've probably already found it. Since m is pretty big, we're probably not going to manage to check as many numbers as we've already checked m times over.

Looking into it more, they've been using more clever ways to prove that it's true up to certain numbers, and we know it's true for up to about 4*10^18. It would be impossible to even check a number that size without some kind of clever method, let alone check enough to find the counter-example.